Convex Hull is Smallest Convex Set containing Set

Theorem

Let $X$ be a vector space over $\R$.

Let $U \subseteq X$ be non-empty.

Let $\map {\operatorname {conv} } U$ be the convex hull of $U$.

Then $\map {\operatorname {conv} } U$ is the smallest convex subset of $X$ containing $U$ in the sense that:

$\map {\operatorname {conv} } U$ is convex and if $K \subseteq X$ is a convex subset with $U \subseteq K$, we have that $\map {\operatorname {conv} } U \subseteq K$.

Corollary

Let $K \subseteq X$ be non-empty.

Then:

$K$ is convex if and only if $\map {\operatorname {conv} } K = K$

Proof

We have:

$\ds \map {\operatorname {conv} } U = \set {\sum_{j \mathop = 1}^n \lambda_j u_j : n \in \N, \, u_j \in U \text { and } \lambda_j \in \R_{> 0} \text { for each } j, \, \sum_{j \mathop = 1}^n \lambda_j = 1}$

Considering $n = 1$ and $\lambda_1 = 1$, we obtain:

$u \in \map {\operatorname {conv} } U$ for each $u \in U$.

So:

$U \subseteq \map {\operatorname {conv} } U$

Now we prove that $\map {\operatorname {conv} } U$ is convex.

Let $u, v \in \map {\operatorname {conv} } U$.

Then there exists $u_1, u_2, \ldots, u_n \in U$, $v_1, v_2, \ldots, v_m$ and $\lambda_1, \lambda_2, \ldots, \lambda_n, \mu_1, \ldots, \mu_m \in \R_{> 0}$ such that:

$\ds u = \sum_{i \mathop = 1}^n \lambda_i u_i$

and:

$\ds v = \sum_{j \mathop = 1}^n \mu_j v_j$

with:

$\ds \sum_{i \mathop = 1}^n \lambda_i = 1$

and:

$\ds \sum_{j \mathop = 1}^n \mu_j = 1$

Now let $t \in \closedint 0 1$.

We have:

$\ds t u + \paren {1 - t} v = \sum_{i \mathop = 1}^n t \lambda_i u_i + \sum_{j \mathop = 1}^m \paren {1 - t} \mu_j \lambda_j$

We have:

\(\ds \sum_{i \mathop = 1}^n t \lambda_i + \sum_{j \mathop = 1}^m \paren {1 - t} \mu_j\)

\(=\)

\(\ds t \sum_{i \mathop = 1}^n \lambda_i + \paren {1 - t} \sum_{j \mathop = 1}^m \mu_j\)

\(\ds \)

\(=\)

\(\ds t + \paren {1 - t}\)

\(\ds \)

\(=\)

\(\ds 1\)

with $t \lambda_i \ge 0$ and $\paren {1 - t} \mu_j \ge 0$ for each $i, j$.

Since we have $u_i \in U$ for each $i$ and $v_j \in U$ for each $j$, we have:

$t u + \paren {1 - t} v \in \map {\mathrm {conv} } U$

So $\map {\mathrm {conv} } U$ is convex.

Lemma

Let $K \subseteq X$ be a convex set with $U \subseteq K$.

Then:

$\map {\operatorname {conv} } U \subseteq K$

$\Box$

Hence the result.

$\blacksquare$

Sources

• 2020: James C. Robinson: Introduction to Functional Analysis ... (previous) ... (next) $21.5$: The Convex Hull